Acids and Bases

Chemistry 12 PACKAGE #5 UNIT 4 ACIDS AND BASES KaKb = Kw Proof: H2O (l) + H2O (l) ⇄ H3O+ (aq) + OH- (aq) Recall, at 25oC: Kw = [H3O+][OH-] [H3O+] = [OH-] = 10-7 M Kw = (10-7) (10-7) = 10-14 Consider the dissociation of any acid: HX (aq) + H2O (l) Ka = [H3O+][X-] ⇄ H3O+(aq) + X- (aq) HX (aq) + [HX] Consider the reaction of a basic anion with water: X- (aq) + H2O (l) ⇄ Kb = [HX][OH-] [X-] OH- (aq) KaKb = [H3O+][X-][HX][OH-] [HX] [X-] KaKb = [H3O+][OH-] = Kw at 25oC KaKb = 10-14 USE KaKb = Kw to solve for Kb if you are given only Ka The acid and base in this equation will be conjugate acid-base pairs example: What is the Kb for NH3 ? NH3 (aq) + H2O (l) ⇄ NH4+ (aq) + OH- (aq) Kb (NH3 ) = Kw Ka (NH4+ ) = 1.0 x 10 -14 5.6 x 10-10 = 1.8 x 10-5 pH of (salt) solutions containing an ion which can act as an acid & an ion which can act as a base NH4CH3COO (s) acidic or basic? - NH4CH3COO (s) à NH4+(aq) + CH3COO (aq) Completely soluble + • Consider the ionization of NH4 in water: NH4+(aq) + H2O (l) ⇄ NH3 (aq) + H3O+ (aq) Ka (NH4+ ) = 5.6 x 10-10 - • Consider the ionization CH3COO in water : CH3COO-(aq) + H2O (l) ⇄ CH3COOH (aq) + OH- (aq) USE KaKb = Kw Kb (CH3COO- ) = Kw Ka (CH3COOH) = 1.0 x 10 -14 1.8 x 10-5 = 5.6 x 10-10 Since Ka = Kb, the pH of the salt NH4CH3COO = 7.0 (neutral) If Ka = Kb then the salt solution would be neutral pH = 7 If Ka > Kb then the salt solution would be acidic pH < 7 If Ka < Kb then the salt solution would be basic pH > 7 KaKb = Kw 1. SAMPLE CALCULATIONS Complete the following chart: ACID Ka conjugate BASE HBr HSO4HNO2 H2CO3 NH4+ HS- 2. 1.3 x 10-13 Consider a solution containing 1.0 M CN - (aq). Find its pH. CN - (aq) OH- (aq) + HOH (l) ⇄ HCN (aq) + Kb + 3. Consider a solution containing 1.0 M CH3COO-(aq). Find its pH and [H3O ]. CH3COO-(aq) + HOH (l) ⇄ HCH3COO(aq) + OH- (aq) 4. Consider a 1.0 M solution with [H3O+] = 2.4 x 10 . Calculate the Kb. -12 ?? (aq) + HOH (l) ⇄ 5. ?? (aq) + OH- (aq) -11 Consider a 0.25 M solution with [H3O+] = 6.3 x 10 . Calculate the Kb. ?? (aq) + HOH (l) ⇄ ?? (aq) + OH- (aq) HYDROLYSIS -a reaction between an ion and water which alters the [OH-] and [H+] in the resulting solution. example: CN- (aq) + HOH (l) ⇄ HCN (aq) + OH- (aq) Note: CN- (aq) hydrolysis produces a basic solution Kh = Kb = [HCN][OH-] [CN-] Typical anion hydrolysis: example: + NH4 (aq) + HOH (l) ⇄ + H3O (aq) + NH3 (aq) + Note: NH4 (aq) hydrolysis produces an acidic solution + Typical cation hydrolysis: Kh = Ka = [H3O ][NH3] + [NH4 ] NOT ALL IONS HYDROLYZE!!! example: Cl- (aq) + HOH (l) ⇄ HCl (aq) + OH- (aq) Kb = Kw / Ka = 10-14 / infinity =0 K <<1 therefore reactants are preferred This hydrolysis reaction does not proceed. HCl is a strong acid and will dissociate 100%. You should be able to recognize immediately why these 6 anions are the Anions incapable of hydrolysis: Cl-, HSO4-, I-, Br -, NO3-. ClO4You should be able to recognize immediately why these cations are the Cations incapable of hydrolysis: +2 +2 +2 +2 Na+, K+, Li +, Cs+, Fr+, Rb+, Sr , Ba , Ca , Mg (we addressed previously the consideration of the concept of solubility in basic hydroxides) HYDROLYSIS OF SALTS: NaF ORIGIN STORY... BUT... since we are being provided with We could write the Arrhenius Acid–Base Neutralization to consider the acid and base that this salt could be formed from. NaOH + HF à H2O + NaF strong weak basic salt base acid the NaF (s) as a salt, it is more relevant to consider the DISSOCIATION of the salt in water. PROOF: NaF (s) à Na+ (aq) + F - (aq) And then we ask ourselves...would each of these ions hydrolyze water, to form their conjugate and alter the [H+] and [OH-] ?? HYDROLYSIS OF IONS in the SALT: Bronsted - Lowry Conjugate Acid-Base pairs Na+ - NO!!! + H2O ß DOES NOT HAPPEN! F- - yes Na + F+ H2O ⇌ HF + OHProof that NaF salt acts as a base when dissolved in solution You could use the Kb value of F - and an ICE grid to calculate the exact pH, if given the concentration of the salt. KCl Bronsted - Lowry Hydrolysis Equations: PROOF: will be basic You can find the Kb of F -: -14 -11 Kb = Kw = 1.0 x 10 = 2.9 x 10 Ka 3.5 x 10-4 DISSOCIATION: PROOF: NH4NO3 pH >7 à So the salt solution HYDROLYSIS OF IONS in the SALT: DISSOCIATION: Bronsted - Lowry Hydrolysis Equations: HYDROLYSIS OF IONS in the SALT: NH4 IO3 DISSOCIATION: PROOF: Bronsted - Lowry Hydrolysis Equations: NH4F HYDROLYSIS OF IONS in the SALT: DISSOCIATION: PROOF: Bronsted - Lowry Hydrolysis Equations: HYDROLYSIS OF IONS in the SALT: NH4CH3COO DISSOCIATION: PROOF: Bronsted - Lowry Hydrolysis Equations: HYDROLYSIS OF IONS in the SALT: SUMMARY: HYDROLYSIS OF SALTS: 1) Neither cation or anion hydrolyze: the salt solution will be neutral 2) Only the cation hydrolyzes: the solution will be acidic 3) Only the anion hydrolyzes: the solution will be basic 4) If both the cation and anion hydrolyze: the solution will be: acidic if Ka > Kb neutral if Ka = Kb basic if Ka < Kb Sample Calculation: DETERMINE THE pH of a 0.060 M NaCH3COO SOLUTION. HYDROLYSIS REVIEW EXERCISE: 1 Find the pH of: a) 0.50 M C6H5OH b) 0.50 M KOH c) 0.50 M H2SO4 d) 0.50 M HBr e) 0.50 M H2CO3 f) 0.50 M NaCN g) 0.50 M NH4I h) 0.50 M K2SO3 2 Give numerical proof to support your prediction of acidic, basic, or neutral character for the following compounds: a) NH4F b) NH4HS c) (NH4)3PO4 3 Will HPO4-2 behave as an acid or a base in water at RTP? Show your proof. HINTS for the HYDROLYSIS WORKSHEET: THIS IS A DECEIVING WORKSHEET. YOU NEED TO KNOW A LOT TO COMPLETE THIS WORKSHEET. SUGGESTION FOR SUCCESS: Make up flash cards of these questions and on the back of the flashcard write: whether it is a strong/weak acid/base, the dissocation equation and whether you will be calculating a Ka or a Kb. Also keep track of how many significant figures you need to have throughout each question. FRONT OF FLASHCARD: a) Find the pH of 0.50 M C6H5OH b) Find the pH of 0.50 M KOH c) Find the pH of 0.50 M H2SO4 d) Find the pH of 0.50 M HBr e) Find the pH of 0.50 M H2CO3 f) Find the pH of 0.50 M NaCN g) Find the pH of 0.50 M NH4I h) Find the pH of 0.50 M K2SO3 BACK OF FLASHCARD: weak acid So use Ka to solve for [H+] C6H5OH ⇄ C6H5O- + H+ strong base therefore 100% dissoc. So no Ka or Kb needed [KOH] = [OH-] strong DIPROTIC acid 100% dissoc [H+] is not appreciably >K2 therefore a 2 stage dissoc IS necessary add [H+]1 + [H+]2 to find final [H+] and then pH strong monoprotic acid therefore one step calc: 100% dissoc. weak DIprotic acid [H+]> K2 therefore ignore second stage use Ka1 to solve for [H+] parent base NaOH: strong base so: Na+ will not hydrolyze parent acid HCN: weak acid so: SINGLE HYDROYSIS: CN- only the ANION solution will be basic; use Kb to solve CN- + HOH <==> HCN + OHHI: strong acid parent (I- does not hydrolyze) NH4OH or NH3: weak base parent SINGLE HYDROYSIS: NH4+ only the CATION solution will be acidic; use Ka to solve NH4+ + HOH <==>H+ + NH3 KOH: strong base base (K+ does not hydrolyze) H2SO3: weak acid parent (you could also look at the HSO3- conj. and come to same conclusion) SINGLE HYDROYSIS: SO3-2 only the ANION solution will be basic Question 2: DETERMINE SINGLE HYDROLYSIS (cation or anion) OR DOUBLE HYDROLYSIS (BOTH cation AND anion). Then use Ka versus Kb to determine acidic / basic character. Question 3: Use Ka of HPO4-2 vs Kb of HPO4-2 to determine if this amphiprotic ion is going to act as acid or base. Note that in question 2b you have to do something similar in addressing the fact that HS- could be an acid OR a base! And also consider the NH4+ ion acting as an acid as well. Highest K value determines it. KaKb = Kw 1. SAMPLE CALCULATIONS Complete the following chart: USE KaKb = Kw ACID Ka conjugate BASE Kb HBr ~infinity Br- 0 HSO4- 1.2 x 10 HNO2 4.6 x 10 H2CO3 4.3 x 10 NH4+ 5.6 x 10 HS- 1.3 x 10 -2 SO4 -2 8.3 x 10-13 -4 NO2- 2.2 x 10-11 -7 HCO3- 2.3 x 10-8 -10 NH3 1.8 x 10-5 -13 S -2 7.7 x 10-2 NOTE THAT FROM THE TOP TO THE BOTTOM OF THE LIST IN THE DATA BOOKLET p. 6 THE ACIDS GO FROM STRONGEST TO WEAK (best to least conductor) THE BASES GO FROM WEAK TO STRONGEST (least to best conductor) Consider a solution containing 1.0 M CN - (aq). Find its pH. IGNORE CATION! It must have been a soluble ionic compound and the cation was a spectator ion! (eg. Alkali ion such as Na+ or perhaps NH4+) That is why they haven’t mentioned it. 2. I C E CN - (aq) + HOH (l) ⇄ 1.0 M X -y X 1.0-y X HCN (aq) + -------+y y OH- (aq) -----+y y Kb = [HCN][OH-] [CN-] -14 -5 Kb= Kw = 1.0 X 10 = 2.0 x 10 -10 Ka 4.9 x 10 -5 2.0 x 10 = y2 1.0-y Assume that y <<1.0 since [CN-] > 103 Kb And therefore the 1.0 – y will round back up to 1.0 -5 y2= (2.0 x 10 )(1.0) y = 4.5 x 10-3 so the assumption is valid (meaning that subtracting y would NOT have changed our answer in the denominator, 1.0 - y, above) -3 [OH-] = y = 4.5 x 10 M pOH = 2.35 (2 sig figs) pH = 11.65 (BASIC!) 3. I C E Consider a solution containing 1.0 M CH3COO-(aq). Find its pH and [H3O+]. IGNORE CATION! CH3COO-(aq) + HOH (l) ⇄ CH3COOH (aq) + OH- (aq) 1.0 M X -------------y X +y +y 1.0-y X y y Kb = [HCH3COO][OH-] [CH3COO-] -14 Kb= Kw = 1.0 X 10 -5 Ka 41.8 x 10 -10 = 5.6 x 10 = y2 Assume that y <<1.0 1.0-y -10 y2= 5.6 x 10 -5 y = 2.4 x 10 assumption valid! y = [OH-] pOH = 4.62 (2 sig figs) pH = 9.38 (BASIC!) 4. Consider a 1.0 M solution with [H3O+] = 2.4 x 10-12. Calculate the Kb. So pH = 11.62 (BASIC!) pOH = 2.38 [OH-] = 4.2 x 10-3 I have represented this value as “y” in the table below I C E ?? (aq) + HOH (l) ⇄ 1.0 M X -y X 1.0-y X Kb = ?? (aq) -------+y y + OH- (aq) -----+y y (4 .2 x 10-3)2 1.0 - 4.2 x 10-3 -5 = 1.8 x 10 If we wanted to figure out which base it was… -14 Ka = Kw = 1.0 x 10 Kb 1.8 x 10-5 Ka = 5.6 x 10 -10 = NH4+ So the base was NH3 !!!!! 5. -11 Consider a 0.25 M solution with [H3O+] = 6.3 x 10 . Calculate the Kb. -11 [H3O+] = 10-10.20 = 6.3 x 10 pH = 10.20 (BASIC!) pOH = 3.80 (2 sig figs) -4 [OH-] = 1.6 x 10 ?? (aq) + HOH (l) I C E 0.25 M -y 0.25-y X X X ⇄ ?? (aq) + OH- (aq) -------+y y -----+y y Where Y represents [OH-] = 1.6 x 10-4 y2 0.25-y = (1.6 x 10-4)2 0.25 – 1.6 x 10-4 = 1.0 x 10-7 Kb = Ka Kb = Kw If Kb = 1.0 x 10-7 Then Ka = 1.0 x 10-7 The identity of the acid with Ka = 1.0 x 10-7 is HSO3So therefore the base with Kb = 1.0 x 10-7 is SO3-2 HYDROLYSIS OF SALTS: For the sake of brevity, the (aq) and (l) were not written in every conjugate acid-base reaction. NaF ORIGIN STORY... BUT... since we are being provided with We could write the Arrhenius Acid–Base Neutralization to consider the acid and base that this salt could be formed from. the NaF (s) as a salt, it is more relevant to consider the DISSOCIATION of the salt in water. NaOH + HF à H2O + NaF strong weak basic salt base acid PROOF: NaF (s) à Na+ (aq) + F - (aq) And then we ask ourselves...would each of these ions hydrolyze water, to form their conjugate and alter the [H+] and [OH-] ?? HYDROLYSIS OF IONS in the SALT: Bronsted - Lowry Conjugate Acid-Base pairs Na+ - NO!!! + H2O ß DOES NOT HAPPEN! F- - yes Na + F+ H2O ⇌ HF + OHProof that NaF salt acts as a base when dissolved in solution You could use the Kb value of F - and an ICE grid to calculate the exact pH, if given the concentration of the salt. KCl pH >7 DISSOCIATION: KCl (s) à K+ (aq) + Cl - (aq) Bronsted - Lowry Hydrolysis Equations: K Cl - will be basic You can find the Kb of F -: -14 -11 Kb = Kw = 1.0 x 10 = 2.9 x 10 Ka 3.5 x 10-4 PROOF: + à So the salt solution + H2O ß DOES NOT HAPPEN! + H2O ß DOES NOT HAPPEN! Neither K+ nor Cl - can hydrolyze to form its conjugate. Therefore the pH of the solution is not affected at all, and stays perfectly neutral. No calculation is required. NH4NO3 K+ - NO!!! Cl - - NO!!! pH = exactly 7.00 DISSOCIATION: NH4NO3 (s) à NH4+ (aq) + NO3 - (aq) PROOF: Bronsted - Lowry Hydrolysis Equations: NH4+ + H2O HYDROLYSIS OF IONS in the SALT: ⇌ NH3 + H3O+ NO3 - + H2O ß DOES NOT HAPPEN! You could use the Ka value of NH4+and an ICE grid to calculate the exact pH, if given the initial salt concentration. HYDROLYSIS OF IONS in the SALT: NH4+ - yes NO3 - - NO!!! pH < 7 due to the NH4+ hydrolysis NH4 IO3 DISSOCIATION: NH4IO3 (s) à NH4+ (aq) + IO3 - (aq) HYDROLYSIS OF IONS in the SALT: PROOF: Bronsted - Lowry Hydrolysis Equations: NH4+ + H2Oà NH3 + H3O+ IO3 - + H2O à HIO3 + OH - NH4+ -10 Ka = 5.6 x 10 -14 Kb = Kw = 5.9 x 10 Ka Since Ka of NH4+ > Kb of IO3 Then the NH4IO3 salt will be an acidic salt. NH4F - yes IO3 - - yes pH < 7 Prove it using ICE grids and [H3O+] and [OH-] DISSOCIATION: NH4F (s) à NH4+ (aq) + F - (aq) HYDROLYSIS OF IONS in the SALT: PROOF: Bronsted - Lowry Hydrolysis Equations: -10 NH4+ -11 F- NH4+ + H2Oà NH3 + H3O+ Ka = 5.6 x 10 F- Kb = 2.9 x 10 ⇌ HF + OH> Kb of F - + H2O Ka of NH4+ This will be an acidic salt, but not nearly as acidic as the previous example, since the K values are relatively similar in magnitude in this case. - yes - yes pH < 7 Prove it using ICE grids and [H3O+] and [OH-] NH4CH3COO DISSOCIATION: NH4CH3COO (s) à NH4+ (aq) + CH3COO - (aq) HYDROLYSIS OF IONS PROOF: in the SALT: Bronsted - Lowry Hydrolysis Equations: NH4+ + H2Oà NH3 + H3O+ CH3COO Kb = Kw = Ka -10 Ka = 5.6 x 10 + H2O ⇌ CH3COOH + OH-14 -10 1.0 x 10 = 5.6 x 10 1.8 x 10-4 Since the Ka of NH4+ = Kb of CH3COO This salt will be exactly neutral. K and ICE grid calculations are not necessary to determine the exact pH. NH4+ - yes CH3COO - - yes pH = exactly 7.00 SUMMARY: HYDROLYSIS OF SALTS: 1) Neither cation or anion hydrolyze: the salt solution will be neutral 2) Only the cation hydrolyzes: the solution will be acidic 3) Only the anion hydrolyzes: the solution will be basic 4) If both the cation and anion hydrolyze: the solution will be: acidic neutral basic if Ka > Kb if Ka = Kb if Ka < Kb Sample Calculation: DETERMINE THE pH of a 0.060 M NaCH3COO SOLUTION. CH3COOH weak acid + NaOH strong base à NaCH3COO basic salt SALT IN SOLUTION: NaCH3COO (aq) à Na + (aq) + CH3COO - + HOH (aq) WHICH ION WILL HYDROLYZE? Na+ - NO CH3COO- yes I C E CH3COO 0.060 M -y 0.060 -y + HOH ⇄ XXX XXX XXX CH3COOH --+y y Kb = [CH3COOH][OH-] [CH3COO-] -14 -10 Kb= Kw = 1.0 X 10 = 5.6 x 10 -5 Ka 1.8 x 10 -10 5.6 x 10 = y2 0.060-y Assume that y <<0.060 since [CH3COO-] > 103 Kb And therefore the 0.060 – y will round back up to 0.060 (0.060) ( 5.6 x 10-10) = y2 -11 3.4 x 10 = y2 -6 y = 5.8 x 10 M= [OH-] pOH = 5.24 pH = 8.76 Basic!!! + OH--+y y HYDROLYSIS REVIEW EXERCISE 1 KEY Find the pH of: a) 0.50 M C6H5OH b) 0.50 M KOH c) 0.50 M H2SO4 d) e) f) g) h) 0.50 M HBr 0.50 M H2CO3 0.50 M NaCN 0.50 M NH4I 0.50 M K2SO3 5.09 or 5.10 13.70 0.29 (final [H+] = .512 M) 0.30 3.33 11.50 or 11.51 4.77 or 4.78 10.35 Acidic Basic Acidic Acidic Acidic Basic Acidic Basic 2 Give numerical proof to support your prediction of acidic, basic, or neutral character for the following compounds: a) NH4F Acidic b) NH4HS Basic c) (NH4)3PO4 Basic 3 Will HPO4-2 behave as an acid or a base in water at RTP? Show your proof. Base